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Question

If X and Y are independent and X follows Bernoulli distribution with probability of success 1/2, while Y follows Bernoulli distribution with probability of success 1/3, find: a) the probability mass function of X+Y b) the cumulative distribution function of X+Y c) the range of X+Y d) the expectation of X+Y e) the variance of X+Y f) the expectation of X⋅Y

Answer

a) P(X + Y = 0) = P(X = 0)P(Y = 0) = (1/2)(2/3) = 1/3 P(X + Y = 1) = P(X = 1)P(Y = 0) + P(X = 0)P(Y = 1) = (1/2)(1/3) + (1/2)(2/3) = 1/2 P(X + Y = 2) = P(X = 1)P(Y = 1) = (1/2)(1/3) = 1/6 b) F(0) = P(X + Y ≤ 0) = P(X = 0, Y = 0) = (1/2)(2/3) = 1/3 F(1) = P(X + Y ≤ 1) = P(X = 0, Y = 0) + P(X = 0, Y = 1) + P(X = 1, Y = 0) = 1/3 + (1/2)(1/3) + (1/2)(2/3) = 5/6 F(2) = P(X + Y ≤ 2) = P(X = 1, Y = 1) + P(X = 0, Y = 1) + P(X = 1, Y = 0) = (1/2)(1/3) + (1/2)(2/3) = 2/3 c) Range of X+Y is {0, 1, 2} d) E[X+Y] = E[X] + E[Y] = 5/6 e) Var[X+Y] = Var[X] + Var[Y] = (1/2)(1/2) + (1/3)(2/3) = 7/18 f) E[XY] = E[X]E[Y] = (1/2)(1/3) = 1/6

a) The probability mass function of X+Y is found by calculating the probability of each possible sum of X and Y. Since X and Y are independent Bernoulli trials, their PMFs are P(X=1)=1/2 and P(Y=1)=1/3 respectively. For X+Y=0, X and Y must both be equal to 0, so P(X + Y = 0) = P(X = 0)P(Y = 0) = (1/2)(2/3) = 1/3. Similarly, for X+Y=1, either X=0 and Y=1 or X=1 and Y=0, whose probabilities are calculated as shown above, which gives P(X+Y=1) = 1/2. Finally, for X+Y=2, X and Y must both be equal to 1, so P(X + Y = 2) = P(X = 1)P(Y = 1) = (1/2)(1/3) = 1/6. b) The cumulative distribution function of X+Y is found by summing the probabilities of X+Y for each value up to a given point. It gives the probability of X+Y being less than or equal to that point. F(0) is the probability that X+Y is less than or equal to 0, which is the same as P(X + Y = 0) = 1/3. Similarly, F(1) is the probability that X+Y is less than or equal to 1, which is the sum of the probabilities of X+Y being 0 and 1. Finally, F(2) is the probability that X+Y is less than or equal to 2, which is the sum of the probabilities of X+Y being 0, 1, and 2. c) The range of X+Y is the set of all possible values that X+Y can take. It is {0, 1, 2}, as these are the only possible sums of X and Y. d) The expectation of X+Y is found by adding the expected values of X and Y, since X and Y are independent. As shown above, E(X)=1/2 and E(Y)=1/3. Therefore, E(X+Y)=1/2+1/3=5/6. e) The variance of X+Y is found by adding the variances of X and Y, since X and Y are independent. As shown above, Var(X)=1/2*(1-1/2)=1/4 and Var(Y)=1/3*(1-1/3)=2/9. Therefore, Var(X+Y)=Var(X)+Var(Y)=1/4+2/9=7/18. f) The expected value of X * Y is found by taking the product of the expected values of X and Y, since X and Y are independent. Therefore, E[XY] = E[X]E[Y] = (1/2)(1/3) = 1/6.